|Interesting Math Problem||Math Problems Home | Home | Send Feedback|
Here is a slightly kludgier variation on the solution:
△ABC ~ △EFC => 5 / (x+y) = h/y
△DCB ~ △EFB => 3 / (x+y) = h/x
From those we get 3x = 5y or x/y = 5/3.
Now we venture off the purists' path.
Since it doesn't matter how far apart the walls are (which is true, but not proven here), we arbitrarliy (and conveniently) set them 5 units apart.
That means △ABC and △EFC are isosceles right triangles, with equal legs.
Then x + y = 5, and y = 3/8 * 5 = 15/8 which is also the value of h.
More generally, if the heights of the walls are A and B,
then the intersection height is (A * B)/(A + B).